Simplify; express your answer in exponential form. Assume $n\neq 0, x\neq 0$. $\dfrac{{(nx^{3})^{-1}}}{{(n^{-4}x^{3})^{-1}}}$
Explanation: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(nx^{3})^{-1} = (n)^{-1}(x^{3})^{-1}}$ On the left, we have ${n}$ to the exponent ${-1}$ . Now ${1 \times -1 = -1}$ , so ${(n)^{-1} = n^{-1}}$ Apply the ideas above to simplify the equation. $\dfrac{{(nx^{3})^{-1}}}{{(n^{-4}x^{3})^{-1}}} = \dfrac{{n^{-1}x^{-3}}}{{n^{4}x^{-3}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{-1}x^{-3}}}{{n^{4}x^{-3}}} = \dfrac{{n^{-1}}}{{n^{4}}} \cdot \dfrac{{x^{-3}}}{{x^{-3}}} = n^{{-1} - {4}} \cdot x^{{-3} - {(-3)}} = n^{-5}$